# Primitive Roots of Unity

**Primitive \(n^\text{th}\) roots of unity** are roots of unity whose multiplicative order is \( n.\) They are the roots of the \(n^\text{th}\) cyclotomic polynomial, and are central in many branches of number theory, especially algebraic number theory.

## Definition

Let \(n\) be a positive integer. A

primitive \(\ n^\text{th}\) root of unityis an \(n^\text{th}\) root of unity that is not a \(k^\text{th}\) root of unity for any positive \( k<n.\) That is, \( \zeta \) is a primitive \(n^\text{th}\) root of unity if and only if\[ \zeta^n=1, \text{and } \zeta^k \ne 1 \text{ for any positive integer } k < n. \]

There are four \( 4^\text{th}\) roots of unity given by \( \pm 1, \pm i.\) Two of these, namely \( \pm i,\) are primitive. The other two are not: \( 1^1 = 1\) and \( (-1)^2=1.\)

## Basic Properties

The primitive \(n^\text{th}\) roots of unity are the complex numbers

\[ e^{2\pi i k/n} : 1\le k \le n, \text{gcd}(k,n)=1. \]

There are \( \phi(n)\) primitive \(n^\text{th}\) roots of unity, where \( \phi(n)\) is Euler's totient function.

Let \( \zeta_n = e^{2\pi i/n}.\) Recall that the \(n^\text{th}\) roots of unity are the \(n\) distinct powers \( \zeta_n^k = e^{2\pi i k/n} : 1 \le k \le n.\) So it remains to show that \( \zeta_n^k\) is primitive if and only if \(k\) and \(n\) are coprime.

The key fact is that \( \zeta_n\) is a primitive \(n^\text{th}\) root of unity, since its first \(n\) powers are distinct. A standard order argument shows that \( \zeta_n^a = 1\) if and only if \( a|n,\) since writing \( n = aq+r,\) \( 0 \le r < n,\) gives \(\zeta_n^r = \zeta_n^{n-aq} = 1,\) but this is impossible unless \( r=0.\)

If \( \text{gcd}(k,n) =d,\) then it is easy to check that \( (\zeta_n^k)^{n/d} = 1,\) so if \( d>1 \) then \( \zeta_n^k\) is not primitive. \(\big(\)In fact, it is not hard to show that \( \zeta_n^k\) is a primitive \(\big(\frac nd\big)^\text{th}\) root of unity.\(\big)\)

If \( \text{gcd}(k,n)=1,\) and \( (\zeta_n^k)^r=1,\) then \(n|kr,\) but \(n\) and \(k\) are coprime so \(n|r.\) This shows that \( \zeta_n^k\) is a primitive \(n^\text{th}\) root of unity, because the first positive power of \(\zeta_n^k\) that equals \( 1\) is \( (\zeta_n^k)^n.\)

Euler's totient function counts the number of positive integers \(k \le n\) that are coprime to \(n,\) which are precisely the exponents that produce primitive \(n^\text{th}\) roots of unity, so this is the number of primitive \(n\)th roots of unity. \(_\square\)

Classify the \(12^\text{th}\) roots of unity by their multiplicative order.

Let \( \zeta_{12} = e^{2\pi i/12}.\) Then the powers of \( \zeta_{12}\) are classified according to the GCD of the exponent and \( 12:\) as in the above proof, \( \zeta_{12}^k\) is a \( \left(\frac{12}{\gcd (12,k)}\right)^\text{th}\) root of unity.

\( \zeta_{12}, \zeta_{12}^5, \zeta_{12}^7, \zeta_{12}^{11} \) are primitive \( 12^\text{th}\) roots of unity. \((\)Note \( \phi(12)=4.)\)

\( \zeta_{12}^2, \zeta_{12}^{10} \) are primitive \(6^\text{th}\) roots of unity.

\( \zeta_{12}^3, \zeta_{12}^9\) are primitive \(4^\text{th}\) roots of unity.

\( \zeta_{12}^4, \zeta_{12}^8\) are primitive \(3^\text{rd}\) roots of unity.

\( \zeta_{12}^6 = -1\) is a primitive \(2^\text{nd}\) root of unity.

\(\zeta_{12}^{12} = 1 \) is a primitive \(1^\text{st}\) root of unity.

Note that the orders are the divisors of \( 12.\) There are \( \phi(d)\) primitive roots of order \( d,\) for each \( d|12.\) Since \( \sum\limits_{d|12} \phi(d) = 12,\) this accounts for all of the \(12^\text{th}\) roots of unity. \(_\square\)

The above arguments show that the powers of a primitive \(n^\text{th}\) root of unity enumerate all the primitive \(d^\text{th}\) roots of unity, for all the divisors \(d\) of \(n.\)

Let \( \zeta_m\) be a primitive \(m^\text{th}\) root of unity, and

let \( \zeta_n\) be a primitive \( n^\text{th}\) root of unity.

Then \( \zeta_m\zeta_n\) is a primitive \(\ell^\text{th}\) root of unity for some positive integer \( \ell.\)

What can we say about \( \ell\) in general?

\(\)

**Clarification:** In the answer choices, \(\gcd(\cdot) \) and \(\text{lcm}(\cdot) \) denote the greatest common divisor function and the lowest common multiple function, respectively.

## Sum and Product

The product of the primitive \(n^\text{th}\) roots of unity is \( 1\) unless \( n=2.\) This is because the set of primitive \(n^\text{th}\) roots of unity, \(n\ge 3,\) can be partitioned into pairs \( \big\{ \zeta^k, \zeta^{n-k} \big\}, \) which multiply to give \( 1.\) \((\)For \( n=2\) this fails because \( \zeta^1 \) and \( \zeta^{2-1} \) coincide.\()\)

The sum of the primitive \(n^\text{th}\) roots of unity is \( \mu(n),\) where \( \mu\) is the Möbius function; see that wiki for a proof.

In fact, there is an elegant formula for the sum of the \(k^\text{th}\) powers of the primitive \(n^\text{th}\) roots of unity:

The sum of the \(k^\text{th}\) powers of the primitive \(n^\text{th}\) roots of unity is

\[ \mu(r) \frac{\phi(n)}{\phi(r)}, \]

where \( r = \frac{n}{\gcd (n,k)}.\)

The extremal examples of the theorem are when \( \text{gcd}(n,k) =1\) and when \( \text{gcd}(n,k)=n.\)

When \( \text{gcd}(n,k)=1,\) taking \(k^\text{th}\) powers permutes the primitive \(n^\text{th}\) roots of unity, so the sum should still be \( \mu(n).\) Indeed, \( r=n,\) so \( \mu(r)\frac{\phi(n)}{\phi(r)} = \mu(n).\)

When \( \text{gcd}(n,k)=n,\) the powers are all \( 1,\) so the sum is \( \phi(n).\) In this case \( r=1,\) so \( \mu(r)\frac{\phi(n)}{\phi(r)} = \phi(n)\) as desired.

Here is an outline: the sum \( \sum \zeta^k \) is a sum of primitive \(r^\text{th}\) roots of unity, and it runs over all of them. But there are repetitions: the sum has \(\phi(n)\) terms and there are \(\phi(r)\) primitive \(r^\text{th}\) roots of unity, so they are each counted \( \frac{\phi(n)}{\phi(r)}\) times. The sum of the primitive \(r^\text{th}\) roots of unity is \(\mu(r),\) so the result follows. \(_\square\)

## Group Structure

The \(n^\text{th}\) roots of unity form a cyclic group under multiplication, generated by \( \zeta_n = e^{2\pi i/n}.\) This group is isomorphic to the additive group \({\mathbb Z}/n\) of the integers mod \(n,\) under the isomorphism \( \zeta_n^k \mapsto k \pmod n.\) In this context, the primitive \(n^\text{th}\) roots of unity correspond via this isomorphism to the other generators of the group. This is the set of multiplicative units \( ({\mathbb Z}/n)^*\) \((\)which has \(\phi(n)\) elements\().\)

The \(12^\text{th}\) roots of unity are isomorphic to \( {\mathbb Z}/12.\) The primitive \(12^\text{th}\) roots of unity, \( \zeta_{12}, \zeta_{12}^5, \zeta_{12}^7,\) and \(\zeta_{12}^{11},\) correspond to the elements \(1,5,7,11\) in \({\mathbb Z}/12,\) respectively. These are the elements which generate \( {\mathbb Z}/12 \) additively. E.g. the multiples of \(5\) are

\[ 5,10,3,8,1,6,11,4,9,2,7,0, \, 5,10,\ldots. \]

Every element in \({\mathbb Z}/12\) is on that list.

## Cyclotomic Polynomials

The polynomial \[ \prod_{\zeta \text{ a primitive } n\text{th root of unity}} (x-\zeta) \] is a polynomial in \(x\) known as the \(n\)th cyclotomic polynomial. It is of great interest in algebraic number theory. For more details and properties, see the wiki on cyclotomic polynomials.

**Cite as:**Primitive Roots of Unity.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/primitive-roots-of-unity/